Calculate the volume (in mL) of 0.100 M H_2O_2 required to oxidise 5.0 g of I_2 completely according to the reaction I_2 + 2 H_2O_2 → 2 I^- + O_2 + 2 H_2O.

Step-by-step solution

1. Convert mass of I_2 to moles: $n_{I_2}=\frac{5.0\text{ g}}{253.8{\text{ g mol}}^{-1}}=0.0197\text{ mol}$.
2. Use stoichiometry: 1 mol I_2 requires 2 mol H_2O_2, so $n_{H_2{O}_2}=2\times 0.0197=0.0394\text{ mol}$.
3. Convert moles of H_2O_2 to volume using concentration: $V=\frac{n}{C}=\frac{0.0394\text{ mol}}{0.100{\text{ mol L}}^{-1}}=0.394\text{ L}$.
4. Express in mL: 0.394 L = 394 mL.
5. Thus the required volume is 394 mL.

Concept tested: Stoichiometry of redox reactions

Published 10 October 2026 · How this solution was prepared and checked